I gave some thoughts about ortogonal matrices and the inverse matrix. I found a interesting point:

Let $A \in \mathbb{R}^2$ with $A=\begin{pmatrix} a & b \\ -c & d \end{pmatrix}$ and $det(A)=1$. The inverse matrix has the form $A^{-1}=\begin{pmatrix} a & -b \\ c & d \end{pmatrix}$. Is it always true, that the entries are the same, except $\pm$?

Thanks, I got it. Quite simple, if I would use the definition...
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ulead86Jan 30 '14 at 21:07

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@ulead86 what do you mean with "only provides $det(A)\neq0$? This formula holds for every invertible $2\times 2$-matrix. If $det(A)=0$ the matrix is not invertible and the question what the entries of $A^{-1}$ look like is meaningless
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user127.0.0.1Jan 30 '14 at 21:12

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If the determinant equals $1$, just change the position of the element on the main diagonal and replace the other two elements by their negatives.
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Michael HoppeJan 30 '14 at 21:12